you haven't shown the graph, but I suppose it would be showing the ...
one and ONLY x intercept! Take that value, and turn it into a factor.
For instance, if the graph intersected at -1, then your factor would be (x- -1) = (x+1).
The given imaginary zeros, in your case 1 + i,ALWAYS come in conjugate PAIRS.
That mean that you have another: 1 -i cool, huh
turn each of those into a factor:
[ x - (1 - i)] = [ x - 1 + i ] and...
[ x - (1 + i)] = [ x - 1 - i ]
regrouping these allows you to get...
[ (x-1) + i ] & [ (x-1) - i ] (**)
you might recognize this as the pattern (a-b) &(a+b) which you can multiply and get... a2 - b2
This coming back...??
so take the terms at (**) and multiply them to get...
[ (x-1)2 - i2 ] which is...
x2 - 2x + 1 + 1 = x2 - 2x + 2 cool. so the two imaginary roots gave us:
(x2 - 2x + 2)
multiply the above by the root you see in your graph (that you didn't show)... for example:
(x - a) (x2 - 2x + 2)
When you are done, you will have a cubic equation...
f(x) = x3 + etc. etc....
that's about it! Let me know how it goes!
The graph of a cubic polynomial function y = f(x) is shown. It is known that one of the zeros is...
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